Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{x^2 - 6x + 8}{x^2 - x - 2} \times \dfrac{x - 9}{3x - 12} $
Answer: First factor out any common factors. $q = \dfrac{x^2 - 6x + 8}{x^2 - x - 2} \times \dfrac{x - 9}{3(x - 4)} $ Then factor the quadratic expressions. $q = \dfrac {(x - 2)(x - 4)} {(x - 2)(x + 1)} \times \dfrac {x - 9} {3(x - 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (x - 2)(x - 4) \times (x - 9)} { (x - 2)(x + 1) \times 3(x - 4)} $ $q = \dfrac {(x - 2)(x - 4)(x - 9)} {3(x - 2)(x + 1)(x - 4)} $ Notice that $(x - 2)$ and $(x - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {\cancel{(x - 2)}(x - 4)(x - 9)} {3\cancel{(x - 2)}(x + 1)(x - 4)} $ We are dividing by $x - 2$ , so $x - 2 \neq 0$ Therefore, $x \neq 2$ $q = \dfrac {\cancel{(x - 2)}\cancel{(x - 4)}(x - 9)} {3\cancel{(x - 2)}(x + 1)\cancel{(x - 4)}} $ We are dividing by $x - 4$ , so $x - 4 \neq 0$ Therefore, $x \neq 4$ $q = \dfrac {x - 9} {3(x + 1)} $ $ q = \dfrac{x - 9}{3(x + 1)}; x \neq 2; x \neq 4 $